If a and b are any two numbers then
(a- b)3= a3-3a2b+ 3ab2- b3
3ab(a-b)= 3a2b- 3ab2

so (a-b)3+ 3ab(a-b)= a3- b3.

In particular, if we let x= a-b, m= 3ab, and n= a3- b3, that says that x3+ mx= n. That is, we can pick any two numbers a, b and right down a cubic equation that has x= a- b as a root.
The question is, can we go the other way: given m and n, can we find a and b so we can write x= a-b as a solution.
The answer to that question is "Yes, we can"!

Since m= 3ab, b= m/3a. Putting that int n= a3- b3, we have [tex]n= a^3- \frac{m^3}{3^3a^3}[/tex].
Multiplying both sides of the equation by a3, we have
[tex]na^3= a^6- (\frac{m}{3})^3[/tex]
which looks worse but is just a quadratic equation in a3:
[tex](a^3)^2- n(a^3)- (\frac{m}{3})^3[/tex].
Use the quadratic formula to solve that
[tex]a^3= \frac{n +/- \sqrt{n^2- 4\left(\frac{m}{3}\right)^3}}{2}[/tex]
[tex]a^3= \frac{n}{2} +/- \sqrt{\left(\frac{n}{2}
\right)^2- \left(\frac{m}{3}\right)^3}[/tex]